Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

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Problem G

Security Panel

Input: standard input

Output: standard output

Time Limit: 2 seconds

Memory Limit: 32 MB

Advanced Control Mechanisms (ACM) produces sophisticated electronic locks and security devices.

The company's most recent invention is a panel of illuminated buttons in r rows and c columns. The buttons are numbered left-to-right, top-to-bottom, starting at 1 in the upper-left corner. Each button has two states: lit and unlit. Initially, all buttons are unlit. Pressing a button switches the state of some buttons from lit to unlit (or vice-versa) according to a 3x3 pattern. Pressing a button on a panel applies the pattern centered on that button. To unlock the panel, the buttons must be pressed in such a way so as to light all of them.

For example, consider the following pattern where pressing a button switches the state of the button pressed, as well as the button above and the buttons to the upper and lower left.

If we use this pattern on a 2x3 panel, then pressing buttons 2, 5, and 6 will light all the buttons. If pressed in that order, the state changes of the panel are: (Light green means lights are off and vice versa)

 

Input

Each input case will begin with the number of rows and columns on the panel, 1 <= r, c <= 5 alone on a line. The next three lines describe how pressing a button will affect the nearby lights. This description consists of a 3x3 character grid, where the character "*" indicates that the light in that position switches state (from lit to unlit or from unlit to lit) while "." means its state remains unchanged.

Input ends with 0 0 alone on a line.

Output

For each input case, output "Case #" followed by the number of the case. If there is no way to turn on all the lights, print "Impossible." If it is possible to turn on the lights, output the buttons to be pressed in increasing order, separated by single space. Output the answer that requires the fewest number of buttons possible to be pressed. If there is more than one correct solution, anyone will do.

Sample Input

2 3

**.

.*.

*..

4 5

.*.

***

.*.

2 2

...

.**

...

4 3

*.*

...

..*

0 0

Sample Output

Case #1

2 5 6

Case #2

2 3 4 7 9 12 14 17 18 19 

Case #3

1 3 

Case #4

Impossible.

(The Decider Contest, Source: University of Alberta Local Contest)

題目類似於關燈問題，但是關燈問題的影響限定十字。

而這題給定影響的情況，問有沒有可能從全關變成全開。

首先最大只有 5x5，考慮將狀態壓縮成一個 int，之後的反轉電燈則使用 XOR 運算即可。

同時預處理每一格壓下的遮罩，純搜索可能會高達 O(2^25)，

依序窮舉時，前兩排的燈必然已經全開了，藉此當作剪枝。

#include <stdio.h>
#include <algorithm>
using namespace std;
int mark[10][10];
int row[10];
int n, m;
int ch[105], found;
void dfs(int x, int y, int state, int step) {
    if(found)   return;
    if(y == m) {
        if(x) {
            if((state&row[x-1]) != row[x-1])
                return;
        }
        x++, y = 0;
    }
    int i, j;
    /*for(i = 0; i < n; i++, puts(""))
        for(j = 0; j < m; j++)
            printf("%d", (state>>(i*5+j))&1);
    printf("-------%d %d %d\n", x, y, step);*/
    if(x == n) {
        if((state&row[x-1]) != row[x-1])
            return;
        found = 1;
        for(i = 0; i < step; i++) {
            if(i)   putchar(' ');
            printf("%d", ch[i]+1);
        }
        puts("");
        return;
    }
    dfs(x, y+1, state, step);
    ch[step] = x*m+y;
    dfs(x, y+1, state^mark[x][y], step+1);
}
int main() {
    /*freopen("in.txt","r+t",stdin);
    freopen("out2.txt","w+t",stdout);*/
    char g[5][5];
    int i, j, k, p, q, r;
    int a, b;
    int cases = 0;
    while(scanf("%d %d", &n, &m) == 2 && n) {
        for(i = 0; i < 3; i++)
            scanf("%s", g[i]);
        int x, y;
        for(i = 0; i < n; i++) {
            for(j = 0; j < m; j++) {
                int &val = mark[i][j];
                val = 0;
                for(p = 0, a = -1; p < 3; p++, a++) {
                    for(q = 0, b = -1; q < 3; q++, b++) {
                        x = i+a, y = j+b;
                        if(x < 0 || y < 0 || x >= n || y >= m)
                            continue;
                        if(g[p][q] == '*')
                            val |= 1<<(x*5+y);
                    }
                }
            }
        }
        /*for(i = 0; i < n; i++, puts(""))
            for(j = 0; j < m; j++)
                printf("%d ", mark[i][j]);*/
        for(i = 0; i < n; i++) {
            row[i] = 0;
            for(j = 0; j < m; j++)
                row[i] |= 1<<(i*5+j);
        }
        found = 0;
        printf("Case #%d\n", ++cases);
        dfs(0, 0, 0, 0);
        if(found == 0)
            puts("Impossible.");
    }
    return 0;
}
/*
2 3
**.
.*.
*..
4 5
.*.
***
.*.
2 2
...
.**
...
4 3
*.*
...
..*
0 0
*/